230433
Elimination of bromine from 2-bromobutane results in the formation of:
1 Predominantly 2-butyne
2 Predominantly l-butene
3 Predominantly 2-butene
4 Equimolar mixture of 1 and 2-butene
Explanation:
:
AIEEE 2005
Hydrocarbons
230434
Reaction of one molecule of $\mathrm{HBr}$ with one molecule of 1,3 -butadiene at $40^{\circ} \mathrm{C}$ gives predominantly:
1 1-bromo-2-butene under kinetically controlled conditions
2 3-bromobutene under thermodynamically controlled conditions
3 1-bromo-2-butene under thermodynamically controlled conditions
4 3-bromobutene under kinetically controlled conditions
Explanation:
: 1, 2-addition product is kinetically controlled product while 1, 4-addition product is thermodynamically controlled product and formed at comparatively higher temperature.
AIEEE 2005
Hydrocarbons
230435
The reaction of propene with $\mathrm{HOCl}\left(\mathrm{Cl}_2+\right.$ $\left.\mathrm{H}_2 \mathrm{O}\right)$ proceeds through the intermediate:
: The reaction of propene with $\mathrm{HOCl}$ proceeds through the intermediate $\mathrm{CH}_3-\stackrel{\oplus}{\mathrm{CH}}-\mathrm{CH}_2 \mathrm{Cl}$.
JEE Main 2016
Hydrocarbons
230484
The presence of $\mathrm{Ag}^{+}$ion increases the solubility of alkenes due to the formation of:
1 $\mathrm{d} \pi$-do bonding
2 $p \sigma-p \pi$ bonding
3 $\mathrm{p} \pi-\mathrm{d} \pi$ bonding
4 $\mathrm{p} \pi-\mathrm{d} \sigma$ bonding
5 non of the above
Explanation:
$\mathrm{Ag}^{+}$forms complex with the alkene by $\mathrm{p} \pi-\mathrm{d} \pi$ bonding giving anion and the solubility is increase.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Hydrocarbons
230433
Elimination of bromine from 2-bromobutane results in the formation of:
1 Predominantly 2-butyne
2 Predominantly l-butene
3 Predominantly 2-butene
4 Equimolar mixture of 1 and 2-butene
Explanation:
:
AIEEE 2005
Hydrocarbons
230434
Reaction of one molecule of $\mathrm{HBr}$ with one molecule of 1,3 -butadiene at $40^{\circ} \mathrm{C}$ gives predominantly:
1 1-bromo-2-butene under kinetically controlled conditions
2 3-bromobutene under thermodynamically controlled conditions
3 1-bromo-2-butene under thermodynamically controlled conditions
4 3-bromobutene under kinetically controlled conditions
Explanation:
: 1, 2-addition product is kinetically controlled product while 1, 4-addition product is thermodynamically controlled product and formed at comparatively higher temperature.
AIEEE 2005
Hydrocarbons
230435
The reaction of propene with $\mathrm{HOCl}\left(\mathrm{Cl}_2+\right.$ $\left.\mathrm{H}_2 \mathrm{O}\right)$ proceeds through the intermediate:
: The reaction of propene with $\mathrm{HOCl}$ proceeds through the intermediate $\mathrm{CH}_3-\stackrel{\oplus}{\mathrm{CH}}-\mathrm{CH}_2 \mathrm{Cl}$.
JEE Main 2016
Hydrocarbons
230484
The presence of $\mathrm{Ag}^{+}$ion increases the solubility of alkenes due to the formation of:
1 $\mathrm{d} \pi$-do bonding
2 $p \sigma-p \pi$ bonding
3 $\mathrm{p} \pi-\mathrm{d} \pi$ bonding
4 $\mathrm{p} \pi-\mathrm{d} \sigma$ bonding
5 non of the above
Explanation:
$\mathrm{Ag}^{+}$forms complex with the alkene by $\mathrm{p} \pi-\mathrm{d} \pi$ bonding giving anion and the solubility is increase.
230433
Elimination of bromine from 2-bromobutane results in the formation of:
1 Predominantly 2-butyne
2 Predominantly l-butene
3 Predominantly 2-butene
4 Equimolar mixture of 1 and 2-butene
Explanation:
:
AIEEE 2005
Hydrocarbons
230434
Reaction of one molecule of $\mathrm{HBr}$ with one molecule of 1,3 -butadiene at $40^{\circ} \mathrm{C}$ gives predominantly:
1 1-bromo-2-butene under kinetically controlled conditions
2 3-bromobutene under thermodynamically controlled conditions
3 1-bromo-2-butene under thermodynamically controlled conditions
4 3-bromobutene under kinetically controlled conditions
Explanation:
: 1, 2-addition product is kinetically controlled product while 1, 4-addition product is thermodynamically controlled product and formed at comparatively higher temperature.
AIEEE 2005
Hydrocarbons
230435
The reaction of propene with $\mathrm{HOCl}\left(\mathrm{Cl}_2+\right.$ $\left.\mathrm{H}_2 \mathrm{O}\right)$ proceeds through the intermediate:
: The reaction of propene with $\mathrm{HOCl}$ proceeds through the intermediate $\mathrm{CH}_3-\stackrel{\oplus}{\mathrm{CH}}-\mathrm{CH}_2 \mathrm{Cl}$.
JEE Main 2016
Hydrocarbons
230484
The presence of $\mathrm{Ag}^{+}$ion increases the solubility of alkenes due to the formation of:
1 $\mathrm{d} \pi$-do bonding
2 $p \sigma-p \pi$ bonding
3 $\mathrm{p} \pi-\mathrm{d} \pi$ bonding
4 $\mathrm{p} \pi-\mathrm{d} \sigma$ bonding
5 non of the above
Explanation:
$\mathrm{Ag}^{+}$forms complex with the alkene by $\mathrm{p} \pi-\mathrm{d} \pi$ bonding giving anion and the solubility is increase.
230433
Elimination of bromine from 2-bromobutane results in the formation of:
1 Predominantly 2-butyne
2 Predominantly l-butene
3 Predominantly 2-butene
4 Equimolar mixture of 1 and 2-butene
Explanation:
:
AIEEE 2005
Hydrocarbons
230434
Reaction of one molecule of $\mathrm{HBr}$ with one molecule of 1,3 -butadiene at $40^{\circ} \mathrm{C}$ gives predominantly:
1 1-bromo-2-butene under kinetically controlled conditions
2 3-bromobutene under thermodynamically controlled conditions
3 1-bromo-2-butene under thermodynamically controlled conditions
4 3-bromobutene under kinetically controlled conditions
Explanation:
: 1, 2-addition product is kinetically controlled product while 1, 4-addition product is thermodynamically controlled product and formed at comparatively higher temperature.
AIEEE 2005
Hydrocarbons
230435
The reaction of propene with $\mathrm{HOCl}\left(\mathrm{Cl}_2+\right.$ $\left.\mathrm{H}_2 \mathrm{O}\right)$ proceeds through the intermediate:
: The reaction of propene with $\mathrm{HOCl}$ proceeds through the intermediate $\mathrm{CH}_3-\stackrel{\oplus}{\mathrm{CH}}-\mathrm{CH}_2 \mathrm{Cl}$.
JEE Main 2016
Hydrocarbons
230484
The presence of $\mathrm{Ag}^{+}$ion increases the solubility of alkenes due to the formation of:
1 $\mathrm{d} \pi$-do bonding
2 $p \sigma-p \pi$ bonding
3 $\mathrm{p} \pi-\mathrm{d} \pi$ bonding
4 $\mathrm{p} \pi-\mathrm{d} \sigma$ bonding
5 non of the above
Explanation:
$\mathrm{Ag}^{+}$forms complex with the alkene by $\mathrm{p} \pi-\mathrm{d} \pi$ bonding giving anion and the solubility is increase.