Question 1: Identify A,B,C,D,E and F in the following series of reaction.
Solution:
Question 2: Give the major product (with proper explanation) when following halogen compounds are treated with sodium ethoxide
Solution:
Note: There can be elimination from 2° and 3° carbonium to give alkene.
Question 3: What are the products of the following reactions?
OCO3 (nucleophile) can't attack 3° carbon having high electron - density hence elimination takes place giving.
Nucleophilic attack on methyl carbon is possible giving ether (Williamson synthesis).
Question 3: Heating many alkyl chlorides or bromides in water effects their conversion into alcohol through SN1 reaction. Order each of the following sets compounds with respect to solvolytic reactivity.
Solution:
(a) I < II < III
(b) III < II < I
(c) III < I < II (allylic)
Order is based on stability of carbocation, formed as the intermediate, which is
1° < 2° < 3° < allylic
Question 5: When alkyl halides are treated with aqueous AgNO3, silver halide precipitate and an alcohol is formed. From what you know about the SN1 reaction, propose a mechanism for the following conversion.
Solution:
Question 6: 2-Bromopentane, when treated with alcoholic KOH yields a mixture of three alkenes A,B and C. Identify A,B and C. Which is predominant?
(Assume reaction proceeds through E2 mechanism)
Solution:
By Saytzeff rule, substituted alkenes are more stable, hence B or C is predominant than A. Out of cis and trans, trans is more stable. Hence C is predominant.
Question 7: Vinyl chloride does not give SN reaction but allyl chloride gives. Explain
Solution: In vinyl chloride C—Cl bond is stable due to resonance (as in chlorobenzene).
Hence SN reaction in which Cl is replaced by nucleophile is not possible. In addition to this, sp2 hybridised carbon is more acidic than sp3 carbon, hence removal of proton (H+) is easier than removal of halide (Cl–).
In allyl chloride, SN reaction is easier since allyl carbonium ion formed after removal of Cl– is tabilised by resonance.
Question 8: When CH3—CH=CH–CH2Cl reacts with alcoholic KCN, a mixture of isomeric products is obtained. Explain.
Solution: It can undergo SN1 and SN2 reaction. By SN2 reaction only one product is formed . But by SN1 reaction, intermediate is carbonium ion.
Thus we get two isomeric products suing SN1 reaction.
Question 9: C4H8Cl2 (A) on hydrolysis forms C4H8O (B) which forms an oxime, but does not reduce Fehling solution. B also gives iodoform test. Identify A and B and explain reactions.
Solution: (A) by hydrolysis forms (C) replacing two —Cl by one O. Thus A has two Cl atoms at same carbon i.e., gem positions.
We assign therefore following structures and compare their properties to match the given properties.
Question 10: Identify end product of the following
Solution:
CBr4 + MeLi → MeBr + LiCBr3
(halogen-metal exchange)
LiCBr3 → LiBr +:CBr2
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