question_answer

Out of the following compounds given, the total number of compound that cannot be prepared by Gabriel Phthalimide synthesis are

I.  t-Butylamine

II. Neopentylamine

III. Diethylamine

IV. p-Toluidine

V. m-Nitroaniline

VI. Vinylamine

VII. p-Nitroaniline

VIII. Allylamine

A) 3                     

B) 4                      

C) 5                      

D)        6

Correct Answer: D

Solution :

[d] Halides or tosylates which are unreactive in \[S{{N}^{2}}\]cannot be used.

(I) to (VI) cannot be prepared but (VII) and (VIII) can be prepared.

(I) It cannot be prepared since \[M{{e}_{3}}C-X\]is \[3{}^\circ \]halide.

(II) It cannot be prepared since \[M{{e}_{3}}C-C{{H}_{2}}-X\]is neopentyl halide and does not undergo \[S{{N}^{2}}\]reaction due to steric hindrance.

(III) It cannot be prepared since \[2{}^\circ \] amine gives poor yield due to elimination.

(IV) It cannot be prepared, since arylhalide, is not activated by EDG (Me group). \[ArSN\]reaction takes place only when the benzene ring is activated by EWG present at o- and p-positions.

(V) It cannot be prepared, since EWG (\[-N{{O}_{2}}\] group)  is present at w-position. \[ArSN\]reaction does not take place.

(VI) It cannot be prepared, since vinyl halides\[(C{{H}_{2}}=\,=CH-X)\] are unreactive in \[S{{N}^{2}}\]reactions

(VII) It can be prepared, since benzene ring is activated by EWG (\[-N{{O}_{2}}\] group) at p-position and hence \[ArSN\] reaction takes place.

(VIII) It can be prepared, since allyl halides\[(C{{H}_{2}}=\,=CHC{{H}_{2}}X)\] can undergo \[S{{N}^{2}}\] reaction.

Note: Aryl halide are unreactive towards \[S{{N}^{2}}\] reaction, due to resonance stabilisation . But, containing EWG (\[\bar{e}\]withdrawing groups)   (e.g., \[-N{{O}_{2}}\]) at o-and p-position activates the benzene ring for \[ArSN\]reaction. That is why o- and p-nitroaniline can be prepared but \[ArX\]containing EDG (\[\bar{e}\]-donating groups) cannot be used for the preparation of corresponding amines.