Answer:
Enolate Alkylation
The anions from ketones, called enolates, can act as a nucleophile in SN2 type reactions. Overall an α hydrogen is replaced with an alkyl group and a new carbon-carbon bond is formed. These alkylations are affected by the same limitations as SN2 reactions previously discussed. A good leaving group, chloride, bromide, iodide or tosylate, should be used. Also, secondary and tertiary leaving groups should not be used because of poor reactivity and possible competition with elimination reactions. Lastly, it is important to use a strong base, such as LDA or sodium amide, for preparing the enolate from the ketone. Using a weaker base such as hydroxide or an alkoxide leaves the possibility of multiple alkylations occurring, and competing SN2 reactions with the base.
Explanation:
Design is illustrated in the attached document
Answer: None of the given options show polymer made up of H₂C=CH-CN (Acrylonitrile).
Explanation: Acrylonitrile (H₂C=CH-CN) which is a monomer on self linkage results in a large chain polymer called as
Polyacrylonitrile.
The structure of Polyacrylonitrile is as follow,
--(H₂C-CHCN-)n--Where n shows the number of Acrylonitrile units joined together in the formation of Polyacrylonitrile. This polymerization reaction can take place by different mechanisms including free radical mechanism, acid catalyzed addition or base catalyzed addition reaction.
The polymerization is shown below,
Answer is: 31,45%.
mrs₁(C₉H₁₆O₄-<span>azelaic acid) = 12g.
mr</span>₂(C₉H₁₆O₄) = 50g.
ω₂(C₉H₁₆O₄) = 15% = 0,15.
mrs₂(C₉H₁₆O₄) = mr₂·ω₂ = 50g·0,15 = 7,5g.
mrs₃(C₉H₁₆O₄) = mrs₁ + mr₂ = 12g + 7,5g = 19,5g.
mr₃ = mr₂ + mr₂ = 50g + 12g = 62g.
ω₃ = mrs₃÷mr₃ = 19,5g ÷ 62g = 31,45% = 0,3145.
The Ions present in CaCl₂ are,
Ca²⁺ Cl⁻ Cl⁻
Means 1 formula unit contains 1 Ca²⁺ ion and 2 Cl⁻ ions.
Also, 1 mole of CaCl₂ contains 6.022 × 10²³ formula units.
So, 1 mole formula units of CaCl₂ contain,
2 × 6.022 × 10²³ = 1.20 × 10²⁴ Cl⁻ Ions
Now, Calculating number of moles contained by 220 g of CaCl₂,
As,
110.98 g of CaCl₂ = 1 mole
Then,
220 g of CaCl₂ = X moles
Solving for X,
X = (220 g × 1 mol) ÷ 110.98 g
X = 1.98 moles
As,
1 mole contained = 1.20 × 10²⁴ Cl⁻ Ions
Then,
1.98 mole will contain = X Cl⁻ Ions
Solving for X,
X = (1.98 mol × 1.20 × 10²⁴ Ions) ÷ 1mol
X = 2.38 × 10²⁴ Cl⁻ Ions