Question

C4H10 (t, kat) → X1 (Br2) → X2 → 1,4-дибромбутан (КОНводн.) → X3 (KMnO4, H2SO4) → X4

Answer 1

(X1)C4H10 (t, kat) → CH2=CH-CH=CH2 +2H2
(X2)CH2=CH-CH=CH2 +(CCl4) Br2---  CH2Br-CH=CH-CH2Br(CCl4)
CH2Br-CH=CH-CH2Br+H2(t, kat)------CH2Br-CH2-CH2-CH2Br
(X3)CH2Br-CH2-CH2-CH2Br+2(КОНводн.)----5CH2(OH)-CH2-CH2CH2(OH)+2KBr+2H2O
(X4)5CH2(OH)-CH2-CH2CH2(OH) + 8KMnO4 + 12H2SO4 --->
-----> 8MnSO4 + 4K2SO4 + 22H2O + 5HOOC-CH2-CH2-COOH

Answer 2

The chemical reaction described in the prompt is a multi-step reaction that involves the conversion of C4H10 (butane) to X4 through several intermediate steps. I'll break down each step of the reaction:

Step 1: C4H10 (butane) reacts with a catalyst (t, kat) to form X1. C4H10 (t, kat) → X1

Step 2: X1 reacts with Br2 (bromine) to form X2. X1 (Br2) → X2

Step 3: X2 reacts with aqueous KOH (potassium hydroxide) to form 1,4-dibromobutane. X2 → 1,4-дибромбутан (КОНводн.)

Step 4: 1,4-dibromobutane reacts with KMnO4 (potassium permanganate) and H2SO4 (sulfuric acid) to form X4. X3 (KMnO4, H2SO4) → X4

Overall, this reaction converts butane (C4H10) to 1,4-dibromobutane through the intermediate formation of X1 and X2. Finally, 1,4-dibromobutane is oxidized to form X4.