Match the List I with List II and select the correct answer using the codes given below the lists? List I (Reaction of L i A l H 4 with acid derivatives) List II (Products) ( p ) C 6 H 5 C H 2 C O O C H 3 1 . P h C H 2 C H 2 N H M e ( q ) 2 . P h C H 2 O H ( r ) PhCOCl 3 . C 6 H 5 C H 2 C H 2 O H + H O C H 3 ( s ) P h C H 2 C O N H M e 4 .

Question

Match the List I with List II and select the correct answer using the codes given below the lists? List I (Reaction of L i A l H 4 with acid derivatives) List II (Products) ( p ) C 6 H 5 C H 2 C O O C H 3 1 . P h C H 2 C H 2 N H M e ( q ) 2 . P h C H 2 O H ( r ) PhCOCl 3 . C 6 H 5 C H 2 C H 2 O H + H O C H 3 ( s ) P h C H 2 C O N H M e 4 ., (p)-3, (q)-4, (r)-1, (s)-2, (p)-3, (q)-4, (r)-2, (s)-1, (p)-1, (q)-2, (r)-3, (s)-4, (p)-2, (q)-3, (r)-1, (s)-4

MARKS App · Accepted Answer

(p) Esters on reduction form alcohols 1 ° is always obtained from acid portion of the esters. C 6 H 5 C H 2 C O O C H 3 → L i A l H 4 C 6 H 5 C H 2 C H 2 O H + H O C H 3 (q) Anhydrides on reduction form two equivalents of - C H 2 O H . (r) P h C O C l → L i A l H 4 P h C H 2 O H + H C l (s) P h C H 2 C O N H M e → L i A l H 4 P h C H 2 C H 2 N H M e

	List I (Reaction of $L i A l H_{4}$ with acid derivatives)		List II (Products)
$(p)$	$C_{6} H_{5} C H_{2} C O O C H_{3}$	$1 .$	$P h C H_{2} C H_{2} N H M e$
$(q)$		$2 .$	$P h C H_{2} O H$
$(r)$	PhCOCl	$3 .$	$C_{6} H_{5} C H_{2} C H_{2} O H + H O C H_{3}$
$(s)$	$P h C H_{2} C O N H M e$	$4 .$

Search any question & find its solution

Looking for more such questions to practice?