$${S_N}2$$  reactions proceed with inversion of configuration. Since the attacking nucleophile is not necessarily the same as that of leaving group, the product cannot be enantiomer of the substrate and thus necessarily will not have opposite optical rotation. Moreover since only one product is obtained, we can not obtain diastereomers.

The ring to which $$- NH$$  group is attached is activated due to the lone pairs on $$N$$ ( $$+M$$  and $$+E$$  effects ) ; while the ring to which $$- C = O$$   is attached is deactivated. Hene, the electrophile would go to the para-position of the activated ring.

The main conditions for geometrical isomerism are
(i) The molecule must have a double bond and the double bond should have restricted rotation
(ii) The two atoms or groups attached to the same carbon atom must be different.
These two conditions are only obeyed by $$Cl - CH = CH - Cl,$$     so it shows geometrical isomerism.

Allylic and benzylic halides show high reactivity towards $${S_N}1$$  reaction. Further, due to greater stabilisation of allyl and benzyl carbocations intermediates by resonance, primary allylic and primary benzylic halides show higher reactivity in $${S_N}1$$  reactions than other simple primary halides.

Hence, it undergoes nucleophilic reaction readily.

The group showing electron-donating effect ( such as $$ - N{H_2}, - OH$$   ) should stabilise the intermediate $$ions,$$  i.e. makes the ring more reactive towards electrophilic substitution than benzene and are called activating group while electron withdrawing group ( such as $$ - Cl,N{O_2}$$   ) increases the positive charge on ring, thus deactivates the ring. Hence, phenol is more readily attacked by an electrophile and is most reactive towards an electroptile .

$$N$$  has no $$d$$ orbital, hence it can’t accommodate more than 8 electrons in its valence shell.

Order of stability of carbanions is $${1^ \circ } > {2^ \circ } > {3^ \circ }.$$

Correct name is 3-methyl-2-butanone.

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In IUPAC nomenclature, double bond is given more preference than triple bond.
$$\mathop {CH}\limits^4 \mathop { \equiv \mathop C\limits^3 - \mathop {CH}\limits^2 = \mathop {C{H_2}}\limits^1 }\limits_{1 - butene\,\, - 3\,\, - yne} $$