This economical West African chicken dish packs a bit of a punch so for those who would prefer it a little milder I suggest you remove the seeds from the chilli.
It has a strong, flavoursome sauce so it’s best served with simple accompaniments like plain boiled rice or couscous and maybe some fried sweet potato.
MALUWE POULET
(Chicken in a spicy sauce)
Serves 4
INGREDIENTS:
1 oz (25g) butter
1 tbsp groundnut oil
4 lb (1.8kg) chicken, cut into 8 portions
2 large cloves garlic, crushed
1 green chilli, finely chopped
1/8 th tsp dried chilli flakes
½ tsp salt
½ tsp black pepper
½ tsp dried rosemary
juice 1 lemon
1 tbsp vinegar
1 tbsp sugar
1 tbsp tomato puree
15 fl.oz (444ml) chicken stock
1 tbsp beurre manié
METHOD:
In a large, lidded frying pan melt the butter over a moderate heat. Fry the chicken, in batches if
necessary, for about 10 minutes, until well browned all over.
Remove from the pan with a slotted spoon and drain on paper towels.
Pour off all but 1 tbsp of fat from the pan. Add the garlic, green chilli, dried chilli flakes, salt and
pepper and fry, stirring, for 1 minute.
Stir in the rosemary, lemon juice, vinegar, sugar, tomato puree and stock. Bring to the boil, stirring constantly.
Return the chicken to the pan, reduce the heat to very low, cover and simmer gently for 35-45
minutes until the chicken is very tender.
Remove the chicken pieces to a warmed serving dish while you finish the sauce.
Drop small pieces of the beurre manié into the sauce and stir until it has thickened and is smooth. Pour over the chicken and serve immediately.
From What’s the recipe today Jim?
