Question

Ma puteți ajuta la doua exerciții din cele cinci va rog?

Answer 1

Răspuns:

2. CH3-CH=CH2 + Br2 ⇒ CH3-CHBr-CH2Br  (A)

CH3-CH=CH2 + Br2 ⇒ CH3-CHBr-CH2Br  (A)CH3-HC*Br- CH2Br (2 izomeri optici = enantiomeri = stereoizomeri)

CH3-CH=CH2 + Br2 ⇒ CH3-CHBr-CH2Br  (A)CH3-HC*Br- CH2Br (2 izomeri optici = enantiomeri = stereoizomeri)Br-CH2-CH2-CH2- Br; CH3-BrCBr-CH3 ; Br-CHBr-CH2-CH3 ;(izomeri de pozitie)

CH3-CH=CH2 + Br2 ⇒ CH3-CHBr-CH2Br  (A)CH3-HC*Br- CH2Br (2 izomeri optici = enantiomeri = stereoizomeri)Br-CH2-CH2-CH2- Br; CH3-BrCBr-CH3 ; Br-CHBr-CH2-CH3 ;(izomeri de pozitie)nA = nBr2 = 0,04·80/160 = 0,02 moli

CH3-CH=CH2 + Br2 ⇒ CH3-CHBr-CH2Br  (A)CH3-HC*Br- CH2Br (2 izomeri optici = enantiomeri = stereoizomeri)Br-CH2-CH2-CH2- Br; CH3-BrCBr-CH3 ; Br-CHBr-CH2-CH3 ;(izomeri de pozitie)nA = nBr2 = 0,04·80/160 = 0,02 molim = 0,02·202 = 4,04g

CH3-CH=CH2 + Br2 ⇒ CH3-CHBr-CH2Br  (A)CH3-HC*Br- CH2Br (2 izomeri optici = enantiomeri = stereoizomeri)Br-CH2-CH2-CH2- Br; CH3-BrCBr-CH3 ; Br-CHBr-CH2-CH3 ;(izomeri de pozitie)nA = nBr2 = 0,04·80/160 = 0,02 molim = 0,02·202 = 4,04g4. nBr = %Br*M/ABr*100 <=> 2 = 86,02*M/8000 <=> 16000 = 86,02M => M = 186g/mol

CH3-CH=CH2 + Br2 ⇒ CH3-CHBr-CH2Br  (A)CH3-HC*Br- CH2Br (2 izomeri optici = enantiomeri = stereoizomeri)Br-CH2-CH2-CH2- Br; CH3-BrCBr-CH3 ; Br-CHBr-CH2-CH3 ;(izomeri de pozitie)nA = nBr2 = 0,04·80/160 = 0,02 molim = 0,02·202 = 4,04g4. nBr = %Br*M/ABr*100 <=> 2 = 86,02*M/8000 <=> 16000 = 86,02M => M = 186g/molCxH2x-2Br2 = 186 <=> CxH2x +158 = 186 <=> 14x = 28 => x=2

CH3-CH=CH2 + Br2 ⇒ CH3-CHBr-CH2Br  (A)CH3-HC*Br- CH2Br (2 izomeri optici = enantiomeri = stereoizomeri)Br-CH2-CH2-CH2- Br; CH3-BrCBr-CH3 ; Br-CHBr-CH2-CH3 ;(izomeri de pozitie)nA = nBr2 = 0,04·80/160 = 0,02 molim = 0,02·202 = 4,04g4. nBr = %Br*M/ABr*100 <=> 2 = 86,02*M/8000 <=> 16000 = 86,02M => M = 186g/molCxH2x-2Br2 = 186 <=> CxH2x +158 = 186 <=> 14x = 28 => x=2F.M=> C2H2Br2

CH3-CH=CH2 + Br2 ⇒ CH3-CHBr-CH2Br  (A)CH3-HC*Br- CH2Br (2 izomeri optici = enantiomeri = stereoizomeri)Br-CH2-CH2-CH2- Br; CH3-BrCBr-CH3 ; Br-CHBr-CH2-CH3 ;(izomeri de pozitie)nA = nBr2 = 0,04·80/160 = 0,02 molim = 0,02·202 = 4,04g4. nBr = %Br*M/ABr*100 <=> 2 = 86,02*M/8000 <=> 16000 = 86,02M => M = 186g/molCxH2x-2Br2 = 186 <=> CxH2x +158 = 186 <=> 14x = 28 => x=2F.M=> C2H2Br2Br Br

Br | |

|HC=CH

|HC=CH1,2-dibromoetena

|HC=CH1,2-dibromoetena

|HC=CH1,2-dibromoetena Br

|HC=CH1,2-dibromoetena Br |

|HC=CH1,2-dibromoetena Br |C=CH₂ - 1,1-dibromoetena

|HC=CH1,2-dibromoetena Br |C=CH₂ - 1,1-dibromoetena |

|HC=CH1,2-dibromoetena Br |C=CH₂ - 1,1-dibromoetena |Br