
In a reaction

Where, M = Molecules, R = Reagent.
M and R are
(a) \[C{H_3}C{H_2}Cl\] and \[NaOH\]
(b) \[C{H_2}Cl - C{H_2}OH\] and aq. \[NaHC{O_3}\]
(c) \[C{H_3}C{H_2}OH\] and \[HCl\]
(d) \[C{H_2} = C{H_2}\] and heat
Answer
162k+ views
Hint: The halohydrin functional group containing a halogen and a hydroxyl group, which are connected to the adjacent carbon atoms. For example: 2-chloroethanol, 3-chloropropane-1,2-diol, etc. Bromohydrin and chlorohydrin are also examples of halohydrin.

Complete Step by Step Answer:
The addition of hypohalous acids (\[HOX\], \[X = Br,Cl\]) to the alkene is an example of an additional reaction. In the given question, we have to determine the molecule (M) and Reagent (R).
As we know that when a molecule of hypohalous acids (\[HOX\], \[X = Br,Cl\]) is added to the double bond of an alkene, the formation of \[C{H_2}Cl - C{H_2}OH\] i.e., formation of M is occurred.
\[C{H_2} = C{H_2} + HOCl \to C{H_2}Cl - C{H_2}OH\]
After that molecule, M undergoes hydrolysis with aqueous sodium bicarbonate (\[NaHC{O_3}\]) OR with reagent R to hydrolyze the \[C - Cl\] bond and form the ethylene glycol (\[C{H_2}OH - C{H_2}OH\]).
\[\begin{array}{l}C{H_2}Cl - C{H_2}OH + aq.NaHC{O_3} \to C{H_2}OH - C{H_2}OH\\\end{array}\]
Like the electrophilic addition reaction of an alkene with bromine, the formation of halohydrin also involves the electrophilic addition reaction, in which anti-addition of chlorine and hydroxyl functional groups occurs with the formation of trans products.
In the question, options (a), (c), and (d) will be the incorrect option because the structure of M is \[C{H_2}Cl - C{H_2}OH\]and reagent R is \[NaHC{O_3}\].
Therefore, we can conclude that option (b) will be the correct answer.
Note: The halohydrins are not formed by the direct addition of hypohalous acid (\[HOX\]), instead the alkene reacts with bromine (\[B{r_2}\]) or chlorine (\[C{l_2}\]) in the presence of water(\[{H_2}O\]). Sodium bicarbonate (\[NaHC{O_3}\]) is also known as baking soda.

Complete Step by Step Answer:
The addition of hypohalous acids (\[HOX\], \[X = Br,Cl\]) to the alkene is an example of an additional reaction. In the given question, we have to determine the molecule (M) and Reagent (R).
As we know that when a molecule of hypohalous acids (\[HOX\], \[X = Br,Cl\]) is added to the double bond of an alkene, the formation of \[C{H_2}Cl - C{H_2}OH\] i.e., formation of M is occurred.
\[C{H_2} = C{H_2} + HOCl \to C{H_2}Cl - C{H_2}OH\]
After that molecule, M undergoes hydrolysis with aqueous sodium bicarbonate (\[NaHC{O_3}\]) OR with reagent R to hydrolyze the \[C - Cl\] bond and form the ethylene glycol (\[C{H_2}OH - C{H_2}OH\]).
\[\begin{array}{l}C{H_2}Cl - C{H_2}OH + aq.NaHC{O_3} \to C{H_2}OH - C{H_2}OH\\\end{array}\]
Like the electrophilic addition reaction of an alkene with bromine, the formation of halohydrin also involves the electrophilic addition reaction, in which anti-addition of chlorine and hydroxyl functional groups occurs with the formation of trans products.
In the question, options (a), (c), and (d) will be the incorrect option because the structure of M is \[C{H_2}Cl - C{H_2}OH\]and reagent R is \[NaHC{O_3}\].
Therefore, we can conclude that option (b) will be the correct answer.
Note: The halohydrins are not formed by the direct addition of hypohalous acid (\[HOX\]), instead the alkene reacts with bromine (\[B{r_2}\]) or chlorine (\[C{l_2}\]) in the presence of water(\[{H_2}O\]). Sodium bicarbonate (\[NaHC{O_3}\]) is also known as baking soda.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main Mock Test Series Class 12 Chemistry for FREE

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Classification of Drugs

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Solutions Class 12 Notes: CBSE Chemistry Chapter 1

NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes

NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry

Electrochemistry Class 12 Notes: CBSE Chemistry Chapter 2
