25. (q-2,6b) qaysi alkin gidratlansa, etanal ch₂cho hosil boladi?
a) pentin-1 b) propin
c) butin-2 d) asetilen

26. (q-2,6b) qaysi modda suv bilan reaktsiyaga kirishib,keton hosil qiladi?
1) atsetilen 2) butin-1 3) butin-2
a) 1,3 b) 2,3
c) 1,2 d) 1

27. (q-2,6b) quyidagi uglevodorodni sistematik nomenklatura boyicha nomlang:
ch₂=ch-ch₂-ch(ch₃)-ch₃
a) 3-metilpenten-1 b) 2-metilpenten-4
c) 4-metilpenten-1 d) 4-metilbuten-1

28. (q-2,6b) quyidagi moddalardan qaysi biri geometrik (sis-trans/z va e) izomeriyaga
ega bola olmaydi?
a) buten-2 b) 1,2-dixloroeten
c) 1-xloropropen d) penten-1

29. (q-2,6b) 2-metilbuten-1 toliq gidrogenlansa, hosil bolgan alkanda nechta birlamchi uglerod
atomi mavjud boladi?
a) 1 b) 3
c) 2 d) 4

30. (q-2,6b) benzol yadrosiga yon zanjirlar 1 va 2 holatda joylashsa bu qanday nomlanadi?
a) orto b) para
c) meta d) izo

Question

25. (q-2,6b) qaysi alkin gidratlansa, etanal ch₂cho hosil boladi?
a) pentin-1  b) propin
c) butin-2  d) asetilen

26. (q-2,6b) qaysi modda suv bilan reaktsiyaga kirishib,keton hosil qiladi?
1) atsetilen  2) butin-1  3) butin-2
a) 1,3  b) 2,3
c) 1,2  d) 1

27. (q-2,6b) quyidagi uglevodorodni sistematik nomenklatura boyicha nomlang:
ch₂=ch-ch₂-ch(ch₃)-ch₃
a) 3-metilpenten-1  b) 2-metilpenten-4
c) 4-metilpenten-1  d) 4-metilbuten-1

28. (q-2,6b) quyidagi moddalardan qaysi biri geometrik (sis-trans/z va e) izomeriyaga
ega bola olmaydi?
a) buten-2  b) 1,2-dixloroeten
c) 1-xloropropen  d) penten-1

29. (q-2,6b) 2-metilbuten-1 toliq gidrogenlansa, hosil bolgan alkanda nechta birlamchi uglerod
atomi mavjud boladi?
a) 1  b) 3
c) 2  d) 4

30. (q-2,6b) benzol yadrosiga yon zanjirlar 1 va 2 holatda joylashsa bu qanday nomlanadi?
a) orto  b) para
c) meta  d) izo

Accepted Answer

### Question 25
# Brief Explanations:
To determine which alkyne forms ethanal ($\ce{CH3CHO}$) upon hydration, we analyze the hydration of alkynes. Hydration of alkynes follows Markovnikov's rule (except for acetylene, which gives acetaldehyde). Propene (alkene) or propyne? Wait, the question is about alkyne. Wait, hydration of propyne: Propyne ($\ce{CH3C\equiv CH}$) upon hydration (with $\ce{H2O}$, $\ce{HgSO4}$, $\ce{H2SO4}$) undergoes keto - enol tautomerism. The enol form (from propyne hydration) tautomerizes to acetone? Wait, no, maybe I made a mistake. Wait, acetylene ($\ce{HC\equiv CH}$) hydration gives acetaldehyde (ethanal). Wait, the options: A) Pent - 1 - yne, B) Propyne, C) But - 2 - yne, D) Acetylene (Asetilen). Wait, hydration of acetylene: $\ce{HC\equiv CH + H2O ->[\ce{HgSO4},\ce{H2SO4}] CH2=CHOH}$ (enol), which tautomerizes to $\ce{CH3CHO}$ (ethanal). So the alkyne is acetylene (Asetilen). But let's check the options. Wait, the question is "Qaysi alkin gidratlansa, etanal $\ce{CH3CHO}$ hosil bo'ladi?" (Which alkyne, when hydrated, forms ethanal?). So acetylene (D) is the answer? Wait, no, maybe propyne? Wait, no, let's re - recall. The hydration of alkynes: terminal alkynes (except acetylene) give methyl ketones, acetylene gives acetaldehyde. So acetylene (D) is the correct answer.
# Answer:
D. Asetilen

### Question 26
# Brief Explanations:
We need to find which substance reacts with water to form a ketone. Let's analyze each substance:
1. **Asetilen (Acetylene)**: Hydration of acetylene gives acetaldehyde (an aldehyde), not a ketone.
2. **But - 1 - yne**: Hydration of but - 1 - yne (terminal alkyne) gives a methyl ketone? Wait, no, but - 1 - yne is a terminal alkyne. Wait, the reaction of alkynes with water (in the presence of $\ce{HgSO4}$ and $\ce{H2SO4}$) follows Markovnikov's addition. For non - terminal alkynes (internal alkynes) like but - 2 - yne: $\ce{CH3C\equiv CCH3 + H2O ->[\ce{HgSO4},\ce{H2SO4}] CH3COCH2CH3}$ (a ketone). For terminal alkynes (like acetylene, but - 1 - yne), acetylene gives acetaldehyde (aldehyde), but - 1 - yne gives a ketone? Wait, no, the general rule is that terminal alkynes (except acetylene) give methyl ketones, acetylene gives acetaldehyde. Wait, let's re - check:
   - Asetilen (1): gives aldehyde (not ketone).
   - Butin - 1 (2): terminal alkyne, hydration gives a ketone? Wait, no, the formula of but - 1 - yne is $\ce{CH3CH2C\equiv CH}$. Hydration: $\ce{CH3CH2C\equiv CH + H2O ->[\ce{HgSO4},\ce{H2SO4}] CH3CH2COCH3}$ (a ketone)? Wait, maybe I was wrong earlier. And butin - 2 (3) is an internal alkyne ($\ce{CH3C\equiv CCH3}$), hydration gives $\ce{CH3COCH2CH3}$ (ketone). So substances 1 (asetilen) gives aldehyde, 2 (butin - 1) gives ketone, 3 (butin - 2) gives ketone. Wait, the options: A) 1,3; B) 2,3; C) 1,2; D) 1. So 2 (butin - 1) and 3 (butin - 2) react with water to form ketones. So the correct option is B.
# Answer:
B. 2,3

### Question 27
# Brief Explanations:
To name the hydrocarbon $\ce{CH2=CH - CH2 - CH(CH3)-CH3}$ using systematic nomenclature, we follow IUPAC rules for alkenes:
1. **Find the longest carbon chain with the double bond**: The longest chain has 5 carbon atoms (pentene).
2. **Number the chain to give the double bond the lowest number**: The double bond is at carbon - 1 (since $\ce{CH2=}$ is at the start).
3. **Identify and number the substituents**: There is a methyl group at carbon - 3 (since the carbon chain is numbered as 1: $\ce{CH2=}$, 2: $\ce{CH - }$, 3: $\ce{CH2 - }$, 4: $\ce{CH(CH3)-}$, 5: $\ce{CH3}$? Wait, no, let's number the chain correctly. The structure is $\ce{CH2=CH - CH2 - CH(CH3)-CH3}$. Let's number from the end closest to the double bond. The double bond is between C1 and C2. Then the carbon atoms are: C1: $\ce{CH2=}$, C2: $\ce{CH - }$, C3: $\ce{CH2 - }$, C4: $\ce{CH(CH3)-}$, C5: $\ce{CH3}$. So the substituent (methyl) is at C3. So the name is 3 - metilpenten - 1 (3 - methylpent - 1 - ene).
# Answer:
A. 3 - metilpenten - 1

### Question 28
# Brief Explanations:
Geometric (cis - trans/Z - E) isomerism in alkenes requires that each carbon of the double bond has two different groups attached. Let's analyze each option:
- **Buten - 2 ($\ce{CH3CH=CHCH3}$)**: Each carbon of the double bond has two different groups ($\ce{CH3}$ and $\ce{H}$ on one, $\ce{CH3}$ and $\ce{H}$ on the other), so it can have geometric isomers.
- **1,2 - dixloroeten ($\ce{ClCH=CHCl}$)**: Each carbon of the double bond has two different groups ($\ce{Cl}$ and $\ce{H}$ on one, $\ce{Cl}$ and $\ce{H}$ on the other), so it can have geometric isomers.
- **1 - xloropropen ($\ce{ClCH=CHCH3}$)**: Each carbon of the double bond has two different groups ($\ce{Cl}$ and $\ce{H}$ on one, $\ce{CH3}$ and $\ce{H}$ on the other), so it can have geometric isomers.
- **Penten - 1 ($\ce{CH2=CH - CH2 - CH2 - CH3}$)**: One of the carbons in the double bond (C1) has two $\ce{H}$ atoms attached. So it cannot have geometric isomers because for geometric isomerism, both carbons of the double bond must have two different substituents.
# Answer:
D. Penten - 1

### Question 29
# Brief Explanations:
First, we need to find the product of the complete hydrogenation of 2 - metilbuten - 1 ($\ce{CH2=C(CH3)-CH2 - CH3}$). Complete hydrogenation (adding $\ce{H2}$ across the double bond) gives 2 - methylbutane ($\ce{CH3CH(CH3)CH2CH3}$). Now, we identify the primary (birlamchi) carbon atoms. A primary carbon atom is a carbon atom attached to only one other carbon atom. In 2 - methylbutane:
- The carbon atoms: C1: $\ce{CH3 - }$ (attached to C2), C2: $\ce{CH(CH3)-}$ (attached to C1, C3, and the methyl group), C3: $\ce{CH2 - }$ (attached to C2 and C4), C4: $\ce{CH3 - }$ (attached to C3), and the methyl group on C2: $\ce{CH3 - }$ (attached to C2). So the primary carbon atoms are C1, C4, and the methyl group on C2? Wait, no, primary carbon is a carbon with only one C - C bond. So in $\ce{CH3CH(CH3)CH2CH3}$:
  - C1: $\ce{CH3 - }$ (bonds: C1 - C2), so primary.
  - C4: $\ce{CH3 - }$ (bonds: C4 - C3), so primary.
  - The methyl group on C2: $\ce{CH3 - }$ (bonds: C - C2), so primary. Wait, that's three primary carbon atoms? Wait, no, let's draw the structure: $\ce{CH3 - CH(CH3)-CH2 - CH3}$. So the carbon atoms:
    - C1: $\ce{CH3}$ (attached to C2) - primary.
    - C2: $\ce{CH}$ (attached to C1, C3, and $\ce{CH3}$) - tertiary.
    - C3: $\ce{CH2}$ (attached to C2 and C4) - secondary.
    - C4: $\ce{CH3}$ (attached to C3) - primary.
    - The $\ce{CH3}$ on C2: $\ce{CH3}$ (attached to C2) - primary. So that's three primary carbon atoms (C1, C4, and the methyl on C2).
# Answer:
B. 3

### Question 30
# Brief Explanations:
In benzene, the position of substituents:
- **Orto (ortho)**: Substituents are on adjacent carbon atoms (1 and 2).
- **Para (para)**: Substituents are on carbon atoms 1 and 4.
- **Meta (meta)**: Substituents are on carbon atoms 1 and 3.
- **Izo**: Not a standard benzene substitution term. Since the substituents are at positions 1 and 2, it is ortho.
# Answer:
A. orto

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Question 25

Answer:

Question 26