Question

An alcohol X(C$_4$H$_{10}$O) on dehydration gave alkene (C$_4$H$_8$) as major product, which on bromination followed by treatment with Y gave alkyne C$_4$H$_6$. Alkyne C$_4$H$_6$ does not react with sodium metal. What are X and Y?

• A. \chemfig{CH_3-CH_2-CH(-[2]OH)-CH_3} ; aq. KOH
• B. \chemfig{CH_3-CH_2-CH(-[2]OH)-CH_3}; (i) alc. KOH (ii) NaNH$_2$
• C. \chemfig{CH_3-CH_2-CH_2-CH_2-OH}; alc. KOH
• D. \chemfig{CH_3-CH_2-CH_2-CH_2-OH}; (i) alc. KOH (ii) NaNH$_2$

Hint:

Dehydration follows Zaitsev's rule. NaNH$_2$ is a strong base used for dehydrohalogenation.

Answer 1

The Correct Option is B

The alcohol X must be butan-2-ol because it gives but-2-ene as the major product upon dehydration (according to Zaitsev's rule). But-2-ene, upon bromination, gives 2,3-dibromobutane. Treatment with a strong base like NaNH$_2$ (Y) results in dehydrohalogenation to form but-2-yne. But-2-yne has no terminal hydrogen, so it does not react with sodium metal. Alcoholic KOH would not give the alkyne directly.