Self Studies
Selfstudy
Selfstudy

Haloalkanes and...

TIME LEFT -
  • Question 1
    1 / -0

    Which of the following sequence of reactions (reagents) can be used for the conversion of $$Ph{CH}_{2}{CH}_{3}$$ into $$PhCH={CH}_{2}$$?

  • Question 2
    1 / -0

    Which of the following sequences would yield m-nitro chlorobenzene $$(Z)$$ from benzene?

  • Question 3
    1 / -0

    Bottles containing $$PhI$$ and $$Ph{CH}_{2}I$$ lost their original labels. They were labelled as $$(A)$$ and $$(B)$$ for testing. $$(A)$$ and $$(B)$$ were separately taken in test tubes and boiled with $$NaOH$$ solutions. The end solution in each tube was made acidic with dilutie $$H{NO}_{3}$$ and some $$Ag{NO}_{3}$$ solution was added. Substance $$(B)$$ gave a yellow precipitate. Which of the following statements is true for this experiment?

  • Question 4
    1 / -0

    The product on monobromination of this compound is :

  • Question 5
    1 / -0

    Aryl halides are less reactive towards nucleophilic substitution reaction as compared to alkyl halides due to:

  • Question 6
    1 / -0

    Directions For Questions

    Comprehension # 3
    Nucleophilic aliphatic substitution reaction is mainly of two types : $$S_{N}1$$ and $$S_{N}2$$. The $$S_{N}1$$ mechanism is a two step process. Reaction velocity of $$S_{N}1$$ reaction depends only on the concentration of the substrate. Since product formation takes place by the formation of carbocation, optically active substrate gives $$(+)$$ and $$(-)$$ forms of the product. In most of the cases the product usually consists of $$5-20\%$$ inverted product and $$80-95\%$$ racemised species. The more stable the carbocation, the greater is the proportion of racemisation. In solvolysis reaction, the more nucleophilic the solvent, the greater is the proportion of inversion.

    ...view full instructions

    Which one of the following compound will give $$S_{N}1$$ reaction predominantly?

  • Question 7
    1 / -0

    Directions For Questions

    Comprehension # 3
    Nucleophilic aliphatic substitution reaction is mainly of two types : $$S_{N}1$$ and $$S_{N}2$$. The $$S_{N}1$$ mechanism is a two step process. Reaction velocity of $$S_{N}1$$ reaction depends only on the concentration of the substrate. Since product formation takes place by the formation of carbocation, optically active substrate gives $$(+)$$ and $$(-)$$ forms of the product. In most of the cases the product usually consists of $$5-20\%$$ inverted product and $$80-95\%$$ racemised species. The more stable the carbocation, the greater is the proportion of racemisation. In solvolysis reaction, the more nucleophilic the solvent, the greater is the proportion of inversion.

    ...view full instructions

    Which of the following compounds will give $$S_{N}1$$ and $$S_{N}2$$ reactions with considerable rate?
    I.$$\;C_6H_5-CH_2-Br$$
    II.$$\;CH_2=CH-CH_2-Br$$
    III.$$\,CH_3-CH(Br)CH_3$$
    Select the correct answer from the codes below below.

  • Question 8
    1 / -0

    Directions For Questions

    Comprehension # 2
    Nucleophilic substitution reactions generally expressed as
    $$Nu^-+R\,-\,L\longrightarrow R-\,Nu+L^-$$
    Where $$Nu^-\longrightarrow Nucleophile\:;\:R-L\longrightarrow substrate\:;\:L\longrightarrow leaving\:group$$
    The best leaving groups are those that become the most stable ions after they depart. Since most leaving groups leave as a negative ion, the best leaving groups are those ions that stabilize a negative charge most effectively. A good leaving group should be
    (a) electron-withdrawing to polarize the carbon
    (b) stable once it has left (not a strong base)
    (c) polarisable- to maitain partial bonding with the carbon in the transition state (both $$S_{N}1$$ and $$S_{N}2$$). This bonding helps to stabilise the transition state and reduces the activation energy.

    ...view full instructions

    $$Cl^-\;\;\;\;\;\;\;\;\;\;\;\;CH_3O^-\;\;\;\;\;\;\;\;\;\;\;\;CH_3S^-\;\;\;\;\;\;\;\;\;\;\;\;I^-$$
    $$(I)\;\;\;\;\;\;\;\;\;\;\;\;(II)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(III)\;\;\;\;\;\;\;\;\;\;\;\;(IV)$$
    The correct order of increasing leaving group capability of above anions:

  • Question 9
    1 / -0

    Directions For Questions

    Comprehension # 2
    Nucleophilic substitution reactions generally expressed as
    $$Nu^-+R\,-\,L\longrightarrow R-\,Nu+L^-$$
    Where $$Nu^-\longrightarrow Nucleophile\:;\:R-L\longrightarrow substrate\:;\:L\longrightarrow leaving\:group$$
    The best leaving groups are those that become the most stable ions after they depart. Since most leaving groups leave as a negative ion, the best leaving groups are those ions that stabilize a negative charge most effectively. A good leaving group should be
    (a) electron-withdrawing to polarize the carbon
    (b) stable once it has left (not a strong base)
    (c) polarisable- to maitain partial bonding with the carbon in the transition state (both $$S_{N}1$$ and $$S_{N}2$$). This bonding helps to stabilise the transition state and reduces the activation energy.

    ...view full instructions

    Among the following which is feasible?

  • Question 10
    1 / -0

    Arrange the following in the increasing order of their ability as a leaving group: $$CF_3SO_3^-,\,CH_3SO_3^-\,and\,CH_3COO^-$$

Submit Test
Self Studies
User
Question Analysis
  • Answered - 0

  • Unanswered - 10

  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10
Submit Test
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now