Question

Answer the following questions based on the given paragraph.

Dehydrohalogenation involves removal of the halogen atom together with a hydrogen atom from a carbon atom adjacent to the one with halogen atom. Alcoholic KOH is used for dehydrohalogenation. According to Saytzeff’s rule, when two alkenes may be formed, the alkene which is most substituted is the major product.

(i) Arrange the following alkyl halides in decreasing order of the rate of $\beta -$elimination reaction with alcoholic KOH.

$CH_{3}CH_{2}CH_{2}Br$ $CH_{3} - \underset{CH_{3}}{\underset{|}{CH}} - CH_{2}Br$

(i) (ii)

$CH_{3} - CH_{2} - \underset{Br}{\underset{|}{CH}} - CH_{3}$

(iii)

(1) (ii) > (iii) > (i) (2) (iii) > (ii) > (i)

(3) (i) > (ii) > (iii) (4) (ii) > (i) > (iii)

(ii) The ease of dehydrohalogenation for different halogens is in the order.

(1) Iodide > bromide> chloride (2) Bromide> iodide> Chloride (3) Chloride> bromide> iodide (4) Iodide > Chloride> bromide.

(iii) What are the products of dehydrohalgenation of 2-iodopentatne?

(1) 2-Pentene (major), 1-pentene (minor) (2) 1-Pentene (major), 2- Pentene (minor) (3) 2-Pentene (50%), 1-Pentene (50%) (4) None of these.

• A. (ii) > (iii) > (i)
• B. (iii) > (ii) > (i)
• C. (i) > (ii) > (iii)
• D. (ii) > (i) > (iii)

Answer 1

Ans.

(i) (2) : The ease of dehydrohalogenation is in the order of tertiary > secondary > primary alkyl halide.

(ii) (1) : Iodide > Bromide > Chloride

(iii) (1) : $CH_{3}CH_{2}CH_{2} - \underset{I}{\underset{|}{C}}H - CH_{3}\overset{\quad alc.KOH\quad}{\rightarrow}$