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Carbon due to its chemical property of catenation, tetravalency and the ability to form stable covalent bonds form hydrocarbon and halo-alkanes easily.

The process of removal of hydrogen and halogen from adjacent carbon atoms of haloalkens is called dehydrohalogenation. NaOEt ( Sodium ethoxide) used in Dehydrohalogenation.  It is a strong base and helps remove a β-hydrogen from the haloalkane.

The chemical reaction for

Reaction:

R–CH2 –CHX–R′+NaOEt   ethanol > R–CH=CHR’+NaX+EtOH

(i) 1-Bromo-1-methylcyclohexane: 

      Br
       |
 CH3–C1–cyclohexane  + NaOEt/EtOH → CH3–C=cyclohexane (double bond at C1-C2) + NaBr + EtOH

Major Product: 1-Methylcyclohexene

(ii) 2-Chloro-2-methylbutane

      CH3
           |
 CH3–C–CH2–CH3   + NaOEt/EtOH → CH3–C=CH–CH3 + NaCl + EtOH
           |                                
          Cl                             CH3

Major Product: 2-Methyl-2-butene (major), 2-Methyl-1-butene (minor)

(iii) 2,2,3-Trimethyl-3-bromopentane

       CH3         CH3
         |            |
 CH3–C–C–CH2–CH3  + NaOEt/EtOH → CH3–C=C–CH2–CH3 + NaBr + EtOH
         |   |                           
        CH3  Br                        CH3

Major Product: 2,2,3-Trimethyl-2-pentene

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Disclaimer: This information is sourced from official website and may vary.

(a) 1-butanol is treated with KI in the presence of H₃PO₄ where the OH is being replaced by the iodine and gives H₂O and KH₂PO₄ as the by-product with 1-iodobutane as the final

(b) The conversion of 1-chlorobutane to 1-iodobutane simply takes place by treating the reactant with KI in the presence of acetone. The iodine from KI normally replaces the chlorine from the reactant and gives 1-iodobutane as the final product with KCl as the by product.

(c) The conversion of but-1-ene to 1-iodobutane takes place according to anti-markovnikoff (positive charged ion H⁺ goes to the carbon which has less number of hydrogens and negative part Br- goes to the carbon which has more number of hydrogens.) i.e. the attack of HBr on but-1-ene in the presence of peroxide gives 1-bromobutane as the intermediate which on treatment with NaI+Acetone gives 1- iodobutane as the final product and NaBr as the by-product.

As per the molecular formula $C_5{H}_{10}$ , one thing can be confirmed; the hydrogen is either cycloalkane or it is an alkene. However, alkenes readily react with chlorine in even dark because of the double bond. Therefore, it is possible that the hydrocarbon is cycloalkane. 

Now, let us consider the reactivity. Cycloalkanes are the saturated hydrocarbons with no double bonds. These do not react with Chlore in dark because such a reaction needs UV light for initiating the process of free radical substitution. Haloalkane and Haloarenes NCERT solutions cover this as well as other questions in further detail.

In bright sunlight, the UV light starts free radical substitution reaction. In cyclopentane, the substitution will result in single monochloro product because all hydrogen atoms are equivalent in the molecule. Therefore, by replacing any hydrogen with chlorine atom will result again in the same product i.e. $C_5{H}_9\mathrm{Cl}.$ This confirms that the hydrocarbon is cyclopentane.

Dipole moment of a molecule depends on the electronegativity difference between atoms bonded covalently and geometry of the molecule (how far the atoms are from each other). Dipole moment is important to understand the polarity of a molecule.

The three dimensional structures of the three compounds along with the direction of dipole moment in each of their bonds are given below:-

CCl₄ being symmetrical has zero dipole moment. In CHCl₃, the resultant of the two C-Cl dipole moments is opposed by the resultant of C-H and C-Cl bonds. Since the dipole Moment of latter resultant is expected to be smaller than the former, CHCl₃ has a finite dipole moment (1.03D)

In CH₂Cl_2, the resultant of two C-Cl dipole moments is reinforced by resultant of two C-H dipoles. Therefore, CH₂Cl₂ (1.62D) has a dipole moment higher than that of CHCl₃. Thus, CH₂Cl₂ has the highest dipole moment.

CH₃CH (Cl)CH (Br)CH₃: 2-Bromo-3-chlorobutane

1. Write the name of side substituent according to the alphabetical order. Example: bromo is written before chloro as B comes before
2. Always use a hyphen {-} between a number and a Since Br is attached to 2 position and Cl is attached to third position i.e. it is written as 2-Bromo and 3-chloro.
3. Lastly, write the name of the longest hydrocarbon chain e. butane of 4 carbons.

CHF₂CBrClF :1-bromo-1-chloro-1,2,2-trifluoroethane

1. While writing the IUPAC name, write the name of side substituent according to the alphabetical Eg: bromo is written before chloro and fluoro as B comes before C and F.
2. Always use a hyphen {-} between a number and a Since Br is attached to first carbon and Cl Is also attached to first carbon i.e. it is written as 1-Bromo and 1-chloro.
3. Since there are 3 F present so it is written as
4. And at the last write the name of the longest hydrocarbon chain e. ethane of 2 carbons.

ClCHC≡CCH₂Br: 1-bromo-4-chlorobut-2-yne

1. While writing the IUPAC name, write the name of side substituent according to the alphabetical That is why bromo is written before chloro as B comes before C.
2. Always use a hyphen {-} between a number and a Since Br is attached to 1 position and Cl isattached to fourthposition i.e. it is written as 1-Bromo and 4-chloro.
3. Also there is a presence of triple bond at the second carbon so the suffix used for triple bond is “yne” along with the position of the bond i.e. but-2-yne

(CCl₃)₃CCl

2- (trichloromethyl)-1,1,1,2,3,3,3-heptachloropropane

Explanation:

1. While writing the IUPAC name, write the name of side substituent according to the alphabetical
2. Always use a hyphen {-} between a number and a
3. Because of the presence of 3 CCl₃ group tri is used as a
4. Since 7 Cl atoms are present in all therefore hepta is used a prefix with the position of chlorine. And at the last write the name of the longest hydrocarbon chain e. propane)

CH₃C (p-ClC₆H₄)₂CH (Br)CH₃

2-bromo-3,3-bis- (4-chlorophenyl)butane

Explanation:

1. While writing the IUPAC name, write the name of side substituent according to the alphabetical
2. Always use a hyphen {-} between a number and a
3. Because of the presence of 2 (p-ClC₆H₂) group bis is used as a prefix instead of di because C₆H₄ contains itself contains numbers 6 &
4. And at the last write the name of the longest hydrocarbon chain e. butane)

(CH₃)₃CCH=CClC₆H₄I-p

1-chloro-1- (4-iodophenyl)-3,3-dimethylbut-1-ene

Explanation:

1. While writing the IUPAC name, write the name of side substituent according to the alphabetical
2. Always use a hyphen {-} between a number and a
3. Because of the presence of 2 methyl group di is used as a
4. And the presence of double bond has the ending ‘ene’ and the presence of double bond at the first position is written as but-1-ene.)