Organic Chemistry Reactions Questions And Answers

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tert-butyl alcohol (alcohol) + HCl (X=halogen)
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Sn1, Cl replaces alcohol group. Forms carbocation. Tert-butyl chloride
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1-heptanol (alcohol) + HBr (X=Halogen)
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Forms 1-bromoheptane, through an Sn2 reaction. no carbocation is formed.
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Alcohol + SOCl2 (thionyl chloride) + pyridine
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Converts the alcohol to an alkyl chloride. This is usually only used for primary and secondary alcohols, because tertiary alcohols are very readily converted to chlorides with hydrogen chloride, since a tertiary carbocation is so stable.
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Methane + Cl2 + heat or ultraviolet rays
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results in chlorination of methane (CH3Cl). Happens through a free radical mechanism.
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tert-butyl alcohol (alcohol) + H2SO4 (acid)
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results in an E1 elimination, forming a mixture of alkenes, the product being 2-methylpropene. Rearrangements of the carbocation can and do occur, via methyl and hydride shifts.
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2-bromo-2-methylbutane + ethanol
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Ethanol is a weak base. 2-bromo-2-methylbutane is a tertiary alkyl halide. Therefore, an E1 elimination reaction occurs through a carbocation intermediate. Tertiary alkyl halides such as this one are so sterically hindered to nucleophilic attack that the presence of any anionic Lewis base favors elimination. Usually substitution predominates over elimination only when anionic Lewis bases are absent. E1 mechanism is followed with tertiary and some secondary alkyl halides, and then only when the base is weak or in low concentration.
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Tert-butyl chloride + sodium methoxide (naOCH3)
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Results in an E2 elimination reaction, forming 2-methyl-1-propene. E2 eliminations occur whenever an alkyl halide (primary, secondary, or tertiary) undergoes elimination in the presence of a strong base.
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alkene + HX (X=halogen)
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results in an alkyl halide, with addition of the halogen in accordance with Markovnikov’s rule (halogen adds to carbon having fewer hydrogens, which allows formation of the most stable carbocation.
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2-methylpropene + Acid (H2SO4) + H2O
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results in hydration of the alkene, with an OH group being added by the 2nd part of the reaction with water. OH group is added in ACCORDAnCE WITH MARKOVnIKOV’S RULE
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1-methylcyclopentene + (1.B2H6, diglyme and 2. H2O2, HO-)
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Results in syn hydration of the alkene, with addition of the OH group opposite of Markovnikov’s rule.
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ethylene + Br2
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results in addition of bromine to the alkene, forming 1,2-dibromoethane. The mechanism is such that in the first step, a bromonium ion is formed. Also, the bromines add anti to each other, resulting in the trans versio, because Br- attacks from the side opposite the C-Br bond of the bromonium ion intermediate.
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Ethylene + Br2 + H2O
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Halogen and a hydroxyl group add to opposite faces of the double bond. A bromonium ion ring is formed, and the bromonium ion ring opens by breaking the bond between bromine and the more substituted carbon. (Therefore, halogen adds to the carbon that has the greater number of hydrogens).
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1-butene + HBr + ROOR + light/heat
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Only with Br is the free radical mechanism followed for addition of hydrogen bromide, and then only in the presence of peroxides (and light or heat). Addition is opposite that of Markovnikov’s rule
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3-ethyl-2-pentene + (1. O3, 2. Zn, H2O)
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This forms two aldehydes, with cleavage happening at the double bond. the two aldehydes formed are acetaldehyde and 3-pentanone.
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Alkene + R’COOOH (peroxy acid)
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results in formation of an epoxide across the double bond. Carboxylic acid (R’COOH) is a byproduct.
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(2-CO2CH3-1-propene) + ROOR
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results in a polymerization reaction. the double bond is broken, and a bond is able to be formed to something else. See exam key 2
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Methyl Bromide + HO-
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results in an Sn2 Reaction (methyl is the most reactive in Sn2 since it is the least crowded), forming Methyl alcohol. The reaction proceeds with inversion of configuration.
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Tert-butyl bromide + H2O
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results in an Sn1 reaction (tertiary is most reactive in Sn1 since it forms the most stable carbocation.) forms Tert-butyl alcohol. Carbocation rearrangements can occur, via either methyl or hydride shifts.
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TsO-
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This is a great leaving group. alkyl p-toluenesulfonates undergo nucleophilic substitution at rates that are even faster than those of alkyl iodides.
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Sodium acetylide + 1-bromobutane in the presence of liquid ammonia (YOU GOT THIS RIGHT, BUT LOOK UP THE MECHAnISM FOR IT)
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Forms 1-hexyne. the acidity of terminal alkynes permits them to be converted to their conjugate bases on treatment with sodium amide. these anions are good nucleophiles and react with methyl and primary alkyl halides to form carbon-carbon bonds. Secondary and tertiary alkyl halides cannot be used, because they yield only elimination products under these conditions.
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Geminal (same carbon) or Vicinal (adjacent carbon) dihalide + 2nanH2 (sodium amide)
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results in an elimination reaction and formation of an alkyne.
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alkyne + 2H2 + metal (Pt, Pd, ni, or Rh)
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results in hydrogenation of alkyne, with H’s being added to the alkyne. Addition is syn. Cis alkenes are intermediates in the hydrogenation of alkynes to alkanes.
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1-ethynylcyclohexanol + H2 + lindlar’s catalyst
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results in formation of an alkyne from an alkene. the lindlar’s catalyst poison the catalyst, making it a poor catalyst for alkene hydrogenation while retaining its ability to catalyze the addition of H2 to the triple bond. ADDITIOn IS SYn.
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3-hexyne + na (sodium) + nH3 (ammonia)
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results in a metal ammonia reduction. partial hydrogenation of the alkyne occurs, resulting in formation of an trans alkene. Anti addition of the hydrogens occurs, so there are two products, itself and it’s enantiomer.
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propyne + 2HBr
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results in addition of hydrogen halides in accordance with markovnikov’s rule to give a geminal dihalide. If only 1 mole of HBr is used, an alkenyl halide results (addition still in accordance with Markovnikov’s rule).
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alkyne + H2O + (H+ and Hg2+ or H2SO4 and HgSO4)
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results in formation of a ketone. Addition is in accordance with Markovnikov’s rule.
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Propyne + 1 mol Cl2. then react the product with 1 mol Cl2 again.
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when the first mol of Cl2 reacts with propyne, anti addition of the Chlorines occur, forming a trans dihaloalkene. Results in formation of 1,1,2,2,-tetrachloropropane.
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1-hexyne + (1. O3 2. H2O)
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cleavage of 1-hexyne occurs at the triple bond, resulting in acids being formed. Pentanoic acid and Carbonic acid are formed in this example. But, the carbanoic acid is unstable and reacts with water to form H2O and CO2. cleavage occurs at tripple bond and is replaced by a double bond to an O and an OH group also (single bond to OH group).
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alkyne + 1. BH3*THF, 2. naOH, H2O2
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Results in a ketone. Addition of borane forms an organoborane. oxidation with basic H2O2 forms an enol. Tautomerization of the enol forms a carbonyl compount. the overall result is addition of H2O to the triple bond.
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benzene + HOnO2 (nitric acid) in H2SO4 (sulphuric acid)
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Results in nitrobenzene. The nitronium cation reacts with the double bond, and then the a water molecule comes and attacks a hydrogen attached to that same carbon, and the electrons from the hydrogen carbon bond go to the sigma bond to form a double bond.
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benzene + HOSO2OH (Sulfuric acid) with heat
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Results in Benzenesulfonic acid being formed (Benzene with an SO2OH group attached).
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Benzene + SO3 in H2SO4 (sulfuric acid)
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Results in Benzenesulfonic acid (benzene with an SO2OH group attached). This method is much faster than simply putting benzene with sulfuric acid in heat.
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Benzene + Br2 (or X2) in FeBr3 (or FeX3)
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forms bromobenzene (or whatever halogen was used). The method is that the Iron(III) bromide combines with Bromine to form a lewis acid / lewis base complex. this complex is a good electrophile, and reacts with benzene, where two of the pi electrons of benzene are used to form a bond to bromine and give a cyclohexadienyl cation intermediate. then, the tetrabromoferrate ion attacks the hydrogen, whose electrons go to form the double bond. The byproducts are HBr and Iron(III) bromide.
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benzene + tert-butyl chloride in AlCl3
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forms tert-butylbenzene, with HCl as a byproduct. tert-butyl chloride and aluminum chloride react to form a tert-butyl cation and a tetrachloroaluminate anion. Then, the tert-butyl cation is attacked by the pi electrons of benzene, and a carbon-carbon bond is formed. Loss of a proton from the cyclohexadienyl cation intermediate yields tert-butylbenzene
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benzene + (CH3)2CHCH2Cl in AlCl3
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results in tert-butylbenzene. A hydride shift occurs in order to avoid a very unstable primary carbocation, and a stable tertiary carbocation forms instead. The hydride shift happens along with the ionization of the carbon-chlorine bond.
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Benzene + cyclohexene in H2SO4
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forms cyclohexylbenzene. (benzene with an attached cyclohexane ring.
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Benzene-(C=O)-CH2CH2CH3 + Zn(Hg) + HCl
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This reaction results in a reduction. the C=O group is replaced with a CH2 group.
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Benzene-(C=O)-CH2CH2CH3 + H2nnH2 and KOH and triethylene glycol
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this reaction results in a reduction. the C=O group is replaced with a CH2 group.
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Toluene + KMnO4 + H2O
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results in the methyl group of Toluene being converted to a COOH group (carboxylic acid group).
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3-chloro-3-methyl-1-butene + H2O + na2CO3
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forms 2-methyl-3-buten-2-ol and 3-methyl-2-buten-1-ol. It is an Sn1 reaction. The two products come from the fact that the H2O attacks the carbocation at both of it’s possible locations. The major contributor is the one that has a tertiary carbocation.
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phenoxide ion + 1-chloro-2-pentene in a solution of acetone and heat.
answer

forms 2-pentenyl phenyl ether, with an Sn2 reaction. Typical Sn2 displacements occur when primary allylic halides react with good nucleophiles. With secondary and tertiary allylic halides or under solvolysis conditions, Sn1 rearrangement products result.
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Propene + Cl2
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Chlorination of propene occurs, with the result being allyl chloride + HCl, with a free radical mechanism.
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1,3 butadiene + HBr, at room temperature
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Results in two products, as a result of 1,2 addition and 1,4 addition. the two products are 3-bromo-1-butene and 1-bromo-2-butene
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cyclodecene + nBS in a solution of CCl4 and heat
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forms 3-bromocyclodecene. alkenes react with nBS to give allylic bromides. substitution occurs by a free radical mechanism.
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Benzene + na, nH3 + CH3OH
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forms 1,4-cyclohexadiene. This is an example of a birch reduction. steps 1 and 3 are single electron transfers from the metal, and steps 2 and 4 are proton transfers from the alcohol.
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benzene + na2Cr2O7 + H20, H2SO4, Heat
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no reaction occurs. there are no hydrogens at the benzylic position, so a reaction does not occur (?)
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nitrotoluene + na2Cr2O7 + H2O, H2SO4, Heat
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p-nitrobenzoic acid is formed. The chromic acid oxidizes (?) the methyl group to a carboxylic acid group.
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2-Chloro-2-phenylpropane + CH3CH2OH
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(p. 449) The conditions chosen favor Sn1 substitution over E2 elimination (weakly basic nucleophile, solvolysis (?)). Therefore, the Cl is replaced by the CH3CH2O group, forming 2-ethoxy-2-phenylpropane.
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p-nitrobenzyl chloride + CH3CO2-na+ + acetic acid
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the result is p-nitrobenzyl acetate (the CH3CO2- group replaces the Chlorine atom). Benzylic halides that are secondary resemble secondary alkyl halides in that they undergo substitution only when the nucleophile is weakly basic (such as CH3CO2-). If the nucleophile is a strong base, such as sodium ethoxide, elimination by the E2 mechanism is faster than substitution.
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2-(m-bromophenyl)-2-butene + H2 + Pt
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Hydrogenation of the side-chain double bond of an alkenylbenzene is much easier than hydrogenation of the aromatic ring and can be achieved with high selectivity, leaving the ring unaffected. the product is 2-(m-bromophenyl)butane. the side chain alkene is hydrogenated.
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tert-butylbenzene + na, nH3 + EtOH
answer

results in a birch reduction. In the presence of na, nH3 and a weak base solvent (such as ethanol), arenes are reduced to nonconjugated dienes. Alkyl-substituted arenes give 1,4-cyclohexadienes in which the alkyl group is a substituent on the double bond. The product of this reaction is 3-tert-butyl-1,4-cyclohexadiene
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styrene (benzene with an ethene group attached) + HBr
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results in 1-phenylethyl bromide. The Br and the H add to the double bond in accordance with Markovnikov’s rule. The aryl group stabilizes the benzylic carbocation and controls the regioselectivity of addition to a double bond involving the benzylic carbon.
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ArH + R(C=O)O(C=O)CH3 + AlCl3 (LOOK THIS UP In BOOK)
answer

results in a ketone Ar(C=O)R and a carboxylic acid R(C=O)OH. Acyl cations generated by treating an acyl chloride or acid anhydride with AlCl3 acylate aromatic rings to yield ketones. The arene must be at least as reactive as a halobenzene. Rearrangements can occur, especially with primary alkyl halides.
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Propyl bromide + Li + diethyl ether
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product formed is propyllithium (CH3CH2CH2Li). lithium metal reacts with organic halides to produce organolithium compounds. the organic halide may be alkyl, alkenyl, or aryl. idodides react most and fluorides least readily. bromides are used most often. suitable solvents include hexane, diethyl ether, and tetrahydrofuran (THF).
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CH3MgI + CH3CH2CH2(C=O)H + 1. diethyl ether + 2. H3O+
answer

results in 2-pentanol. Grignard reagents react with formaldehyde to yield primary alcohols, with aldehydes to give secondary alcohols, and with ketones to form tertiary alcohols.
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cyclopropyllithium + 3,3-dimethyl-2-butanone + 1. diethyl ether + 2. H3O+
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product is 2-cyclopropyl-3,3-dimethyl-2-butanol. organolithium reagents react with aldehydes and ketones in a manner similar to that of grignard reagents to produce alcohols.
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sodium acetylide + 2-butanone + 1. nH3 + 2. H3O+
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3-methyl-1-pentyn-3-ol
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(CH3)2CuLi + C6H5CH2Cl + diethyl ether
answer

C6H5CH2CH3. two alkyl groups may be coupled together to form an alkane by th ereaction of an alkyl halide with a lithium dialkylcuprate. both alkyl groups must be PRIMARY OR METHYL. Aryl and vinyl halides may be used in place of alkyl halides.
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cyclopentene + CH2I2, Zn(Cu) + diethyl ether
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results in bicyclo[3.1.0]hexane. Methylene transfer from iodomethylzinc iodide converts alkenes to cyclopropanes. the reaction is a stereospecific syn addition of a CH2 group to the double bond. stereochemistry must remain the same (?)
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propene + 1. BH3, THF + 2. H2O2, naOH
answer

Propyl alcohol is formed. OH adds to the double bond opposite of Markovnikov’s rule. Addition of the H and OH is syn.
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1-cyclobutane-2-ethene + HCl
answer

methylcyclopentyl chloride. adding HCl can break open the ring and expand it.

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