311.
Arrange the carbanions, $${\left( {C{H_3}} \right)_3}\overline C ,\overline C C{l_3},{\left( {C{H_3}} \right)_2}\overline C H,{C_6}{H_5}\overline C {H_2}$$ in order of their decreasing stability :
A
$${\left( {C{H_3}} \right)_2}\overline C H > \overline C C{l_3} > {C_6}{H_5}\overline C {H_2} > {\left( {C{H_3}} \right)_3}\overline C $$
B
$$\overline C C{l_3} > {C_6}{H_5}\overline C {H_2} > {\left( {C{H_3}} \right)_2}\overline C H > {\left( {C{H_3}} \right)_3}\overline C $$
C
$${\left( {C{H_3}} \right)_3}\overline C > {\left( {C{H_3}} \right)_2}\overline C H > {C_6}{H_5}\overline C {H_2} > \overline C C{l_3}$$
D
$${C_6}{H_5}\overline C {H_2} > \overline C C{l_3} > {\left( {C{H_3}} \right)_3}\overline C > {\left( {C{H_3}} \right)_2}\overline C H$$
Answer :
$$\overline C C{l_3} > {C_6}{H_5}\overline C {H_2} > {\left( {C{H_3}} \right)_2}\overline C H > {\left( {C{H_3}} \right)_3}\overline C $$
312.
$$C{H_3}Br + N{u^ - } \to C{H_3} - Nu + B{r^ - }$$ The decreasing order of the rate of the above reaction with nucleophiles $$\left( {N{u^ - }} \right)$$ A to D is
$$\left[ {N{u^ - } = \left( {\text{A}} \right)Ph{O^ - },\left( {\text{B}} \right)Ac{O^ - },\left( {\text{C}} \right)H{O^ - },\left( {\text{D}} \right)C{H_3}{O^ - }} \right]$$
TIPS/Formulae :
The stronger the acid, the weaker the conjugate base formed.
The acid character follows the order :
$$C{H_3}COOH > {C_6}{H_5}OH > {H_2}O > C{H_3}OH$$
The basic character will follow the order
$$C{H_3}CO{O^ - } < {C_6}{H_5}{O^ - } < {O^ - }H < C{H_3}{O^ - }$$
313.
The stability of carbanions in the following
is in the order of
The carbanion with more $$s$$ - character is more stable. Thus, the order of stability is
$$RC \equiv \mathop C\limits^\Theta > {C_6}H_5^\Theta > {R_2}C = \mathop C\limits^\Theta H > {R_3}C - \mathop C\limits^\Theta {H_2}$$
314.
In the given reaction two products are expected.
The product $$(B)$$ is formed as a major product because
A
the carbocation $$C{H_3} - \mathop C\limits^ + H - C{H_3}$$ is formed which is more stable
B
the carbocation $$C{H_3} - C{H_2} - \mathop C\limits^ + {H_2}$$ is formed which is more stable
C
both carbocations are equally stable but the nucleophile attacks on central $$C$$ atom
D
$$C{H_3} - \mathop C\limits^ + H - C{H_3}$$ can easily give a proton to attach $$B{r^ + }.$$
Answer :
the carbocation $$C{H_3} - \mathop C\limits^ + H - C{H_3}$$ is formed which is more stable
$${C_2}{H_5}SH + \frac{9}{2}{O_2}\left( g \right) \to 2C{O_2}\left( g \right) + 3{H_2}O\left( l \right) + S{O_2}\left( g \right)$$
At $$298\,K,C{O_2}\,{\text{and}}\,S{O_2}$$ exist as gases while $${H_2}O$$ exists as liquid.
316.
Which of the following represents the given mode of hybridisation $$s{p^2} - s{p^2} - sp - sp$$ from left to right?
As $$Cl$$ is most electronegative element among $$Cl, Mg, Br$$ and $$C$$ thus, it has greater $$-I$$ effect and the asterisk marked carbon in $$\mathop C\limits^ * {H_3} - C{H_2} - Cl$$ is expected to have greater positive charge.
320.
How many primary, secondary, tertiary and quaternary carbon atoms are present in the following compound?
A
One primary, two secondary and one tertiary
B
Five primary, three secondary
C
Five primary, one secondary, one tertiary and one quaternary
D
Four primary, two secondary and two quaternary
Answer :
Five primary, one secondary, one tertiary and one quaternary