Question

Column I		Column II
A		p	KI + Acetone
B		q	$Zn + CH_3COOH$
C		r	$EtOK + EtOH$
D		s
E		t	$NaOH + CH_3OH$

• A. A -(r), B -(s), C - (q), D - (e), E - (p)
• B. A -(s), B -(r,t), C - (r,t), D - (s), E - (p)
• C. A -(s,r), B -(r,q), C - (e), D - (r), E - (p)
• D. A -(p), B -(r), C - (q), D - (t), E - (p)

Answer 1

Answer: (B)

$(a \rightarrow s )$ Hofmann elimination (less-substituted alkene), so bulky base (s) is required
$( b \rightarrow r , t )$ Saytzeff elimination (more-substituted alkene), so less bulky base ( $r$ and $t$ ) are required
$(c \rightarrow r , t$ ) Same explanation as in (b)
$(d \rightarrow s )$ Same explanation as in (a)
$( e \rightarrow p , q )$ Anti- dehalogenation can be carried out by (p) or reduction of (R-X) to (R-H) by reagent (q)