2022年Edexcel ALevel化学AS方程式总结

根据Ofqual最新的官方数据显示,ALevel化学荣登2020最受欢迎Alevel科目榜单第四!近两年得A/A*率表现也有所提升,2019年得A率29.1% (A*率:7.6%),2020年得A率42.9%(A*率:14.8%)。

2022年1月 ALevel化学考试时间

第一单元:1月12日
第二单元:1月17日
第三单元:1月24日
第四单元:1月7日
第五单元:1月14日
第六单元:1月20日

距离第一门化学考试还剩3天!今天带大家复习AS化学,第一单元、第二单元涉及到的化学反应方程式。

Unit 1

▄Alkanes – with O2
Complete combustion:CH4 + 2O2 → CO2 + 2H2OIncomplete
combustion:CH4 + 1.5O2 →CO + 2H2O

▄ Alkanes – with Cl2, UV
CH4 + Cl2 → CH3Cl + HCl
Mechanism:Photochemical free radical substitution.

▄ Alkanes – improving the quality of fuels
Catalytic Cracking:C10H22 → C5H12 + C5H10       Conditions – Heat (600oC), Al2O3

▄ Alkenes – with H2
H2C=CH2 + H2 →CH3CH3 Conditions - Heat, Ni

▄ Alkenes – with Br2
CH3CH=CH2 + Br2→CH3CHBrCH2Br  Colour change (orange to colourless)
Mechanism:Electrophilic Addition

▄ Alkenes – with Bromine water
CH3CH=CH2 + Br2 / H2O →CH3CHOHCH2BrColour change (orange to colourless)
Mechanism:Electrophilic Addition

▄ Alkenes – with HBr/dry/gas
CH3CH=CH2 + HBr →CH3CH2BrCH3     Major product
Mechanism :Electrophilic Addition
Explanation for major product:Secondary carbocationic intermediate is more stable than primary.

▄ Alkenes – with KMnO4 / H2SO4
CH3CH=CH2 + [O] →CH3CHOHCH2OHColour change (purple to colourless – H2SO4)Alkenes- Polymerisation

Unit 2

▄ Group 2 metals – with H2O
Mg:Mg + H2O → MgO + H2 (steam only)Ca,Sr,
Ba:Ca + 2H2O→ Ca(OH)2 + H2

▄ Group 2 metals – with O2
2Mg + O2 →2MgO

▄ Group 2 metals – with Cl2
Mg + Cl2 →MgCl2

▄ Group 2 Oxides – with H2O
MgO + H2O →Mg(OH)2

▄ Group 2 Oxides – with acids
MgO + H2SO4→ MgSO4 + H2O

▄ Thermal stability
❶ Group 1 Carbonates:All stable to heat except for Li2CO3
❷ Group 2 Carbonates:MgCO3→ MgO + CO2
❸ Group 1 Nitrates Li:   2LiNO3 → Li2O + 2NO2 + ½O2   Na, K, Rb, Cs :NaNO3→NaNO2 + ½O2
❹ Group 2 – Nitrates   Mg(NO3)2→ MgO + 2NO2 + ½O2

▄ Solubility
Group 1 and 2 Sulphates Decreases down the group - BaSO4 is insolublGroup 1 and 2 Hydroxides Increases down the group - Mg(OH)2 is insoluble ▄ Group 7 elements – with H2O
Cl2 + H2O→HCl + HClO
▄ Group 7 elements – with NaOH
In Cold Dilute Alkali: Cl2 + 2OH- →Cl- + ClO- + H2O  Reaction type - Disproportionation In Hot Conc.
Alkali: 3Cl2 + 6OH- → ClO3- + 5Cl- + 3H2O Reaction type - Disproportionation

▄ Group 7 – Displacement reactions
Cl2(g) + 2Br-(aq) → Br2(aq) + 2Cl-(aq)
Observations:Green gas → orange solution

▄ Halide ions – with conc H2SO4:
❶ NaCl(s) + H2SO4 →NaHSO4(s) + HCl(g)
Observations:Steamy white fumes
❷ NaBr(s) + H2SO4→NaHSO4(s) + HBr(g)  2HBr(g) + H2SO4 → Br2(g) + SO2(g) + 2H2O(l)
Observations Steamy white fumes and orange fumes
❸ NaI(s) + H2SO4→Products HI(g), I2(g) + H2S(g)
NaI(s) + H2SO4→NaHSO4(s)+HI(g)
Followed by oxidation of HI(g):2HI(g)+H2SO4→I2(g) + SO2(g)+2H2O(l)
And:6HI(g)+ H2SO4→3I2(g)+S(s)+4H2O(l)
And:8HI(g)+ H2SO4→4I2(g)+ H2S(g) +4H2O(l)
Observations:Steamy white fumes and purple fumes

▄ Test for halide ions
Ag+(aq) + Cl-(aq) →AgCl(s) Observations: White ppt – soluble in dilute ammonia
▄ Halogenoalkanes - with aqueous OH-  CH3CH2Br + OH-→ CH3CH2OH + Br-
Mechanism : Nucleophilic substitution (Sn1 or Sn2)
▄ Halogenoalkanes with ethanolic OH-
CH3CH2Br + OH-→H2C=CH2 + Br- + H2O
Mechanism: Elimination

▄ Halogenoalkanes - with CN-
CH3CH2Br + CN-→CH3CH2CN + Br-
Mechanism: Nucleophilic substitution

▄ Halogenoalkanes - with aqueous silver nitrate
CH3CH2Br + H2O + Ag+→ CH3CH2OH + AgBrFastest
halogenoalkane :Iodo
Explanation: C-I bond is weaker than C-Br and C-Cl

▄ Halogenoalkanes - with NH3
CH3CH2Br + NH3→CH3CH2NH2 + HBrConditions: Conc NH3 / heat / closed vessel▄ Preparation of halogenoalkanes
①Chloroalkanes from alcohols :
CH3CH2OH + Cl- → CH3CH2Cl + OH-
Conditions: H2SO4 / NaCl / heat

②Bromoalkanes from alcohols:
CH3CH2OH + Br-→CH3CH2Br + OH-
Conditions: NaBr / H3PO4 / Heat Not H2SO4 / NaBr / heat as Br2 will form

③Iodoalkanes from alcohols:
CH3CH2OH + I- → CH3CH2I + OH-
Conditions: PI3 or P / I2Not H2SO4 / NaI / heat as I2 will form

▄ Alcohols – 1o Partial Oxidation:
CH3CH2OH + [O] →CH3CHO + H2O
Conditions: Distil product as it is formed

▄ Alcohols – 1o Complete Oxidation:
CH3CH2OH + 2[O] →CH3CO2H + H2O
Conditions: Heating under reflux

▄ Alcohols – 2o Oxidation:
CH3CHOHCH3 + [O] →CH3COCH3 + H2O

▄ Alcohols – Dehydration:
CH3CH2OH→ H2C=CH2 + H2O
Conditions: Heating under reflux / NaOH – Aqueous

▄ Alcohols – Reaction with sodium:
CH3CH2OH + Na →CH3CH2O-Na+ + ½ H2
Observation: Colourless effervescence

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