




Step 1: Elimination Reaction with Alcoholic KOH
When 1-bromopropane (CH3CH2CH2Br) is heated with alcoholic KOH, it undergoes β-elimination to form propene (CH3CH=CH2).
Reaction:
CH3CH2CH2Br + alc. KOH → CH3CH=CH2 + H2O + KBr
So, A is: CH3CH=CH2 (Propene)
Step 2: Electrophilic Addition with Br2
Propene reacts with bromine to give 1,2-dibromopropane.
Reaction:
CH3CH=CH2 + Br2 → CH3CHBrCH2Br
So, B is: CH3CHBrCH2Br (1,2-Dibromopropane)
Final Answer:
| LIST I | LIST II | ||
|---|---|---|---|
| A | Lyman | I | Near IR |
| B | Balmer | II | Far IR |
| C | Paschen | III | Visible |
| D | p-fund | IV | UV |


Match the following:
In the following, \( [x] \) denotes the greatest integer less than or equal to \( x \). 
Choose the correct answer from the options given below:
For x < 0:
f(x) = ex + ax
For x ≥ 0:
f(x) = b(x - 1)2